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QCD is the best-known example of theories with negtive beta function, i.e., coupling constant decreases when increasing energy scale. I have two questions about it:

(1) Are there other theories with this property? (non-Abelian gauge theory, principal chiral field, non-linear sigma model, Kondo effect, and ???)

(2) Are there any simple (maybe deep) reason why these theories are different from others? It seems that the non-linear constraint of the non-linear sigma model (and principal chiral model) is important, but I have no idea how to generalize this argument to other theories.

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Do you want to consider (extended) SUSY or not? –  Vibert May 25 '13 at 7:14

3 Answers 3

As gih correctly pointed out, there are a lot of QFTs in low dimensions which are asymptotically free.

In 3d, for example:

  1. Scalar $\phi^4$ field theory in 3d is asymptotically free. Renormalization flow increases the quartic coupling, taking you away from the free theory.
  2. Likewise, the 3d Gross-Neveu model (Dirac fermions with a 4-fermion interaction) is also asymptotically free. The 4-fermion interaction is relevant in 3d (rather than non-renormalizable, as it is in 4d).
  3. Plain old 3d Yang-Mills gauge theory in 3d is asymptotically free. (Even better, you can set things up so that the YM kinetic term itself is irrelevant, giving you a topological theory at low energies.)

It's not hard to combine these basic models to get more complicated but still asymptotically free models.

To address the 2nd part of your question: There's probably not anything model-specific going on here. It's just that the basic ingredients in these lower dimensional theories are better behaved.

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I disagree with you, there is a deep reason why one can find a RG trajectory in, for example, the $\phi^4$ theory in $d<4$ : it is because you assume that the flow start on a very specific trajectory that link the Gaussian fixed point and the Wilson-Fisher fixed point. If you pick an initial condition randomly, it won't be free in the UV. See also my answer here (and the reference) : physics.stackexchange.com/questions/73403/… –  Adam Jan 16 at 17:04
    
@Adam I understood Tengen to be asking whether there were 'geometrical-ish' features of the models he mentioned which explained why the renormalization trajectories existed. You're right, however, that my second to last paragraph is confusing. –  user1504 Jan 16 at 20:13
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Sure. I just wanted to stress that even the theory that we usually think as asymptotically free are like that only on very specific RG trajectories. –  Adam Jan 16 at 20:21

One example would be $\varphi^3$ theory, which is treated extensively in Srednicki.

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Is it artificial? Since there is no true ground state for $\phi^3$ thoery. The strong coupling property at low energy may indicates the collapse of perturbation theory. –  Tengen Jan 24 '13 at 8:23
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You are right, the theory is unstable. –  Frederic Brünner Jan 24 '13 at 10:21
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@Tengen: A purely $\phi^3$ theory is unstable, but one usually start with a stable theory (for instance, with both $\phi^3$ and $\phi^4$). But when one is using perturbation theory, the $\phi^4$ term is irrelevant compare to the $\phi^3$ and one can forget about it in the calculations. But of course, at the non-perturbative level (at the level of the functional integral), there are always terms that make the theory physical. –  Adam Jan 16 at 20:19

Yes, such examples are generic in 2+1 and 1+1D.

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Could you elaborate on this? You only answer the first part of the question. –  Manishearth Dec 17 '12 at 8:50

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